# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 302: 17

The remaining angles and sides are $$\angle C=57.36^\circ\hspace{.75cm}b\approx11.119ft\hspace{.75cm}c\approx11.544ft$$

#### Work Step by Step

$$\angle A=68.41^\circ\hspace{.75cm}\angle B=54.23^\circ\hspace{.75cm}a=12.75ft$$ 1) Analysis: - Angles $\angle A$ and $\angle B$ are known. We can always calculate $\angle C$ as the sum of 3 angles in a triangle equals $180^\circ$. - Side $a$ and its opposite $\angle A$ are known. $\angle B$ and $\angle C$ are also known, so they are helpful to finding out sides $b$ and $c$. (Law of sines is to be applied) 2) Calculate the unknown angle $\angle C$ We know that the sum of 3 angles in a triangle is $180^\circ$. $$\angle A+\angle B+\angle C=180^\circ$$ $$68.41^\circ+54.23^\circ+\angle C=180^\circ$$ $$122.64^\circ+\angle C=180^\circ$$ $$\angle C=180^\circ-122.64^\circ=57.36^\circ$$ 3) Calculate the unknown sides $b$ and $c$ a) For $b$ We know the opposite angle of $b$: $\angle B=54.23^\circ$, so $\sin B=\sin54.23^\circ\approx0.811$. We also know side $a=12.75ft$ and its opposite angle $\angle A=68.41^\circ$, $\sin A\approx0.93$. Therefore, using the law of sines: $$\frac{b}{\sin B}=\frac{a}{\sin A}$$ $$b=\frac{a\sin B}{\sin A}$$ $$b=\frac{12.75ft\times0.811}{0.93}$$ $$b\approx11.119ft$$ b) For $c$ We know the opposite angle of $c$: $\angle C=57.36^\circ$, so $\sin C=\sin57.36^\circ\approx0.842$. We also know side $a=12.75ft$ and its opposite angle $\angle A=68.41^\circ$, $\sin A\approx0.93$. Therefore, using the law of sines: $$\frac{c}{\sin C}=\frac{a}{\sin A}$$ $$c=\frac{a\sin C}{\sin A}$$ $$c=\frac{12.75ft\times0.842}{0.93}$$ $$c\approx11.544ft$$ 4) Conclusion: The remaining angles and sides are $$\angle C=57.36^\circ\hspace{.75cm}b\approx11.119ft\hspace{.75cm}c\approx11.544ft$$

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